ode primer

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Chapter 08.01Primer for Ordinary Differential Equations

After reading this chapter, you should be able to:

1. define an ordinary differential equation,2. differentiate between an ordinary and partial differential equation, and3.

solve linear ordinary differential equations with fixed constants by using classicalsolution and Laplace transform techniques.

Introduction

An equation that consists of derivatives is called a differential equation. Differential

equations have applications in all areas of science and engineering. Mathematical

formulation of most of the physical and engineering problems leads to differential equations.So, it is important for engineers and scientists to know how to set up differential equations

and solve them.

Differential equations are of two types

(A) ordinary differential equations (ODE)(B) partial differential equations (PDE)An ordinary differential equation is that in which all the derivatives are with respect to asingle independent variable. Examples of ordinary differential equations include

022

2

=++ ydx

dy

dx

yd, 4)0(,2)0( == y

dx

dy,

,sin532

2

3

3

xydx

dy

dx

yd

dx

yd=+++ ,12)0(

2

2

=dx

yd2)0( =

dx

dy, 4)0( =y

Ordinary differential equations are classified in terms of order and degree. Order of an

ordinary differential equation is the same as the highest derivative and the degree of anordinary differential equation is the power of highest derivative.

Thus the differential equation,

xexydx

dyx

dx

ydx

dx

ydx =+++

2

22

3

33

08.01.1

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08.01.2 Chapter 08.01

is of order 3 and degree 1, whereas the differential equation

xdx

dyx

dx

dysin1 2

2

=+

+

is of order 1 and degree 2.An engineers approach to differential equations is different from a mathematician. While,

the latter is interested in the mathematical solution, an engineer should be able to interpret the

result physically. So, an engineers approach can be divided into three phases:a) formulation of a differential equation from a given physical situation,b) solving the differential equation and evaluating the constants, using given conditions,

andc) interpreting the results physically for implementation.

Formulation of differential equations

As discussed above, the formulation of a differential equation is based on a given physical

situation. This can be illustrated by a spring-mass-damper system.

Above is the schematic diagram of a spring-mass-damper system. A block is suspendedfreely using a spring. As most physical systems involve some kind of damping - viscous

damping, dry damping, magnetic damping, etc., a damper or dashpot is attached to account

for viscous damping.

Kb

Figure 1 Spring-mass damper system.

M

Let the mass of the block be , the spring constant be K, and the dampercoefficient be b . If we measure displacement from the static equilibrium position we need

not consider gravitational force as it is balanced by tension in the spring at equilibrium.

Below is the free body diagram of the block at static and dynamic equilibrium. So,the equation of motion is given by

(1)DS FFMa +=

where

is the restoring force due to spring.SF

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Primer for Ordinary Differential Equation 08.01.3

is the damping force due to the damper.DF

is the acceleration.a

The restoring force in the spring is given by

(2)KxFS =

as the restoring force is proportional to displacement and it is negative as it opposes the

motion. The damping force in the damper is given by(3)bvFD =

as the damping force is directly proportional to velocity and also opposes motion.

Therefore, the equation of motion can be written as

(4)bvKxMa =

Since

SF

Ma

DynamicStatic

Mg

T DF

Figure 2 Free body diagram of spring-mass-damper system.

2

2

dt

xda = and

dt

dxv =

from Equation (4), we get

dt

dx

bKxdt

xd

M=

2

2

02

2

=++ Kxdt

dxb

dt

xdM (5)

This is an ordinary differential equation of second order and of degree one.

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08.01.4 Chapter 08.01

Solution to linear ordinary differential equations

In this section we discuss two techniques used to solve ordinary differential equations

(A) Classical technique(B) Laplace transform technique

Classical Technique

The general form of a linear ordinary differential equation with constant coefficients is givenby

)(......... 122

2

31

1

xFykdx

dyk

dx

ydk

dx

ydk

dx

ydn

n

nn

n

=+++++

(6)

The general solution contains two parts

(7)PH yyy +=

where

is the homogeneous part of the solution andHy

is the particular part of the solution.Py

The homogeneous part of the solution is that part of the solution that gives zero when

substituted in the left hand side of the equation. So, is solution of the equation

Hy

Hy

0......... 122

2

31

1

=+++++

ykdx

dyk

dx

ydk

dx

ydk

dx

ydn

n

nn

n

(8)

The above equation can be symbolically written as

(9)0................. 121 =++++ ykDykyDkyD nn

n

(10)0).................( 121 =++++ ykDkDkD nn

n

where,

n

nn

dx

d

D = (11)

.

.

.

1

11

=

n

nn

dx

dD

operating on is the same as

,),( 1rD )( 2rD )( nrD

operating one after the other in any order, where)(....,),........(),( 21 nrDrDrD

are factors of

0 (12)............... 121 =++++ kDkDkD nn

n

To illustrate

0)23(2 =+ yDD

is same as

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0)1)(2( = yDD

0)2)(1( = yDD

Therefore,

(13)0)....................( 121 =++++ ykDkDkD nn

n

is same as

0).........().........)(( 11 = yrDrDrD nn (14)

operating one after the other in any order.

Case 1: Roots are real and distinct

The entire left hand side becomes zero if ( ) 01 = yrD . Therefore, the solution tois a solution to a homogeneous equation.( ) 01 = yrD ( ) 01 = yrD is called Leibnitzs

linear differential equation of first order and its solution is

(15)( ) 01 = yrD

yr

dx

dy1= (16)

dxry

dy1= (17)

Integrating both sides we get

(18)cxry += 1ln

(19)xrcey 1=

Since any of the factors can be placed before , there are different solutions

corresponding to different factors given by

n y n

n

xrxrxr

n

xr

n eCeCeCeCnn 121

121 ,.....,,.........,

where

are the roots of Equation (12) and121, ,..,,......... rrrr nn

are constants.121, ,,......, CCCC nn

We get the general solution for a homogeneous equation by superimposing the individual

Leibnitzs solutions. Therefore

(20)xr

n

xr

n

xrxr

Hnn eCeCeCeCy ++++=

121

121 .............

Case 2: Roots are real and identical

If two roots of a homogeneous equation are equal, say 21 rr = , then

0))(...(..........).........)(( 111 = yrDrDrDrD nn (21)

Lets work at

(22)0))(( 11 = yrDrD

If

(23)zyrD = )( 1

then

0)( 1 = zrD

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08.01.6 Chapter 08.01

(24)xreCz 12=

Now substituting the solution from Equation (24) in Equation (23)xreCyrD 121)( =

xreCyr

dx

dy1

21 =

2111 Cyer

dx

dye

xrxr =

2

)( 1C

dx

yedxr

=

(25)dxCyedxr

2)(1 =

Integrating both sides of Equation (25), we get

121 CxCyexr +=

(26)xr

eCxCy 1)( 12 +=

Therefore the final homogeneous solution is given by

(27)( ) xrnxrxrHneCeCexCCy ++++= ...

31321

Similarly, if m roots are equal the solution is given by

( ) xrnxrmxrmmH nmm eCeCexCxCxCCy +++++++= ++ .......... 1112321 (28)

Case 3: Roots are complex

If one pair of roots is complex, say ir +=1 and ir =2 ,

where

1=i then

(29)( ) ( )xr

n

xrxixi

H

neCeCeCeCy ++++= + ......3321

Since

, and (30a)xixe xi sincos +=

(30b)xixe xi sincos =

then

( ) ( ) xrnxrxx

HneCeCxixeCxixeCy +++++= .........sincossincos 3321

( ) ( ) xrnxrxx neCeCxeCCixeCC +++++= .........sincos 332121

(31)( ) xrnxrx neCeCxBxAe ++++= ........sincos 33

where

and21

CCA +=

(32))( 21 CCiB =

Now, let us look at how the particular part of the solution is found. Consider the general

form of the ordinary differential equation

( ) XykDkDkD nnnnn =++++ 1211 .......... (33)The particular part of the solution is that part of solution that givesPy X when substituted

for in the above equation, that is,

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( ) XykDkDkD Pnnnnn =++++ 1211 ...... (34)

Sample Case 1

When , the particular part of the solution is of the form . We can findaxeX = axAe A by

substituting in the left hand side of the differential equation and equating

coefficients.

axAey =

Example 1

Solve

xeydx

dy =+ 23 , 5)0( =y

Solution

The homogeneous solution for the above equation is given by

( ) 023 =+ yD

The characteristic equation for the above equation is given by023 =+r

The solution to the equation is

666667.0=r

xH Cey666667.0=

The particular part of the solution is of the form xAe

( ) xxx eAedx

Aed

=+ 23

xxx eAeAe =+ 23xx eAe =

1=AHence the particular part of the solution is

x

P ey=

The complete solution is given by

PH yyy +=

xx eCe = 666667.0

The constant can be obtained by using the initial conditionC 5)0( =y

( ) 50 00666667.0 == eCey 51=C 6=C

The complete solution isxx eey = 666667.06

Example 2Solve

xeydx

dy 5.132 =+ , 5)0( =y

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08.01.8 Chapter 08.01

Solution

The homogeneous solution for the above equation is given by

( ) 032 =+ yDThe characteristic equation for the above equation is given by

032 =+rThe solution to the equation is

5.1=r

xH Cey5.1=

Based on the forcing function of the ordinary differential equations, the particular part of the

solution is of the form , but since that is part of the form of the homogeneous part of

the solution, we need to choose the next independent solution, that is,

xAe 5.1

xP Axey5.1=

To find A , we substitute this solution in the ordinary differential equation as

( ) xxx eAxedx

Axed 5.15.15.1

32

=+

xxxx eAxeAxeAe 5.15.15.15.1 332 =+

xx eAe 5.15.12 =

5.0=AHence the particular part of the solution is

x

P xey5.15.0 =

The complete solution is given by

PH yyy +=

xx xeCe 5.15.1 5.0 +=The constant is obtained by using the initial conditionC 5)0( =y .

( ) 5)0(5.00)0(5.1)0(5.1

=+=

eCey 50 =+C

5=CThe complete solution is

xx xeey 5.15.1 5.05 +=

Sample Case 2

When

or ,)sin(axX= )cos(ax

)

the particular part of the solution is of the form

.cos()sin( axBaxA +

We can get andA B by substituting )cos()sin( axBaxAy += in the left hand side of the

differential equation and equating coefficients.

Example 3

Solve

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xydx

dy

dx

ydsin125.332

2

2

=++ , 3)0(,5)0( === xdx

dyy

Solution

The homogeneous equation is given by

0)125.332(

2

=++ yDDThe characteristic equation is

0125.332 2 =++ rrThe roots of the characteristic equation are

22

125.32433 2

=r

4

2593

=

4

163

=

4

43

i=

i= 75.0Therefore the homogeneous part of the solution is given by

)sincos( 2175.0

xKxKey xH +=

The particular part of the solution is of the form

xBxAyP cossin +=

( ) ( ) xxBxAxBxAdx

dxBxA

dx

dsin)cossin(125.3cossin3cossin2

2

2

=+++++

( ) xxBxAxBxAxBxAdxd

sin)cossin(125.3)sincos(3sincos2 =+++

xxBxAxBxAxBxA sin)cossin(125.3)sincos(3)cossin(2 =+++

xxABxBA sincos)3125.1(sin)3125.1( =++

Equating coefficients of andxsin cos on both sides, we get

13125.1 = BA03125.1 =+ AB

Solving the above two simultaneous linear equations we get

109589.0=A292237.0=B

Hence

xxyP cos292237.0sin109589.0 =The complete solution is given by

)cos292237.0sin109589.0()sincos( 2175.0

xxxKxKey x ++=

To find and we use the initial conditions1K 2K

3)0(,5)0( === xdx

dyy

From we get5)0( =y

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08.01.10 Chapter 08.01

))0cos(292237.0)0sin(109589.0())0sin()0cos((5 21)0(75.0 ++= KKe

292237.05 1 = K

292237.51 =K

xx

xKxKexKxKe

dx

dy xx

sin292237.0cos109589.0

)cossin()sincos(75.0 2175.0

21

75.0

++

+++=

From

,3)0( ==xdx

dy

we get

)0sin(292237.0)0cos(109589.0

))0cos()0sin(())0sin()0cos((75.03 21)0(75.0

21

)0(75.0

++

+++= KKeKKe

109589.075.03 21 ++= KK

109589.0)292237.5(75.03 2 ++= K

859588.62 =K

The complete solution is

xxxxey x cos292237.0sin109589.0)sin859588.6cos292237.5(75.0 ++=

Example 4

Solve

)cos(125.3622

2

xydx

dy

dx

yd=++ , 3)0(,5)0( === x

dx

dyy

Solution

The homogeneous part of the equations is given by0)125.362( 2 =++ yDD

The characteristic equation is given by

0125.362 2 =++ rr

)2(2

)125.3)(2(4662

=r

4

25366 =

4

116 =

829156.05.1 =329156.2,670844.0 =

Therefore, the homogeneous solution is given byHy

xxH eKeKy329156.2

2

670845.0

1

+=


xBxAyP cossin +=

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Substituting the particular part of the solution in the differential equation,

xxBxA

xBxAdx

dxBxA

dx

d

cos)cossin(125.3

)cossin(6)cossin(22

2

=++

+++

xxBxA

xBxAxBxAdx

d

cos)cossin(125.3

)sincos(6)sincos(2

=++

+

xxBxA

xBxAxBxA

cos)cossin(125.3

)sincos(6)cossin(2

=++

+

xxABxBA coscos)6125.1(sin)6125.1( =++

Equating coefficients of xcos and we getxsin

06125.1

16125.1

=

=+

BA

AB

The solution to the above two simultaneous linear equations are

0301887.0161006.0

==

BA

Hence the particular part of the solution is

xxyP cos0301887.0sin161006.0 +=

Therefore the complete solution is

PH yyy +=

xxeKeKy xx cos0301887.0sin161006.0)(329156.2

2

670845.0

1 +++=

Constants and can be determined using initial conditions. From ,1K 2K 5)0( =y

50301887.0)0( 21 =++= KKy

969811.40301887.0521 ==+ KK

Now

x

eKeKdx

dy xx

sin0301887.0cos161006.0

329156.2670845.0 )329156.2(2)670845.0(

1

+

=

From 3)0( ==xdx

dy

3161006.0329156.2670845.0 21 =+ KK

161006.03329156.2670845.0 21 +=+ KK

838994.2329156.2670845.0 21 =+ KK

We have two linear equations with two unknowns969811.421 =+ KK

838994.2329156.2670845.0 21 =+ KK

Solving the above two simultaneous linear equations, we get

692253.81 =K

722442.32 =K


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08.01.12 Chapter 08.01

.cos0301887.0sin161006.0

)722442.3692253.8( 329156.2670845.0

xx

eey xx

++

=

Sample Case 3When

or ,bxeX ax sin= bxeax costhe particular part of the solution is of the form

)cossin( bxBbxAeax + ,

we can get andA Bby substituting

)cossin( bxBbxAey ax +=

in the left hand side of differential equation and equating coefficients.

Example 5

Solve

xeydxdy

dxyd x sin125.352 2

2

=++ , 3)0(,5)0( === xdxdyy

Solution

The homogeneous equation is given by

0)125.352( 2 =++ yDD

The characteristic equation is given by

0125.352 2 =++ rr

)2(2

)125.3)(2(455 2 =r

425255 =

4

05 =

25.1,25.1 =

Since roots are repeated, the homogeneous solution is given byHy

xH exKKy)25.1(

21 )(+=


)cossin( xBxAey xP +=

Substituting the particular part of the solution in the ordinary differential equation

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xexBxAe

xBxAedx

dxBxAe

dx

d

xx

xx

sin)}cossin({125.3

)}cossin({5)}cossin({22

2

=++

+++

xexBxAexBxAexBxAe

xBxAexBxAedx

d

xxxx

xx

sin)cossin(125.3)}sincos()cossin({5

)}sincos()cossin({2

=+++++

++

xexBxAexBxAexBxAe

xBxAexBxAexBxAexBxAe

xxxx

xxxx

sin)cossin(125.3)}sincos()cossin({5

)}cossin()sincos()sincos()cossin({2

=+++++

++

xexBxAexBxAe xxx sin)sincos()cossin(875.1 =++

xxBxAxBxA sin)sincos()cossin(875.1 =++

xxBAxBA sincos)875.1(sin)875.1( =++

Equating coefficients of xcos and on both sides we getxsin

0875.1 = BA 1875.1 =+ BA

Solving the above two simultaneous linear equations we get

and415224.0=A221453.0=B

Hence,

)

)

cos221453.0sin415224.0( xxey xP +=

Therefore complete solution is given by

PH yyy +=

cos221453.0sin415224.0()( 25.121 xxeexKKyxx ++=

Constants and can be determined using initial conditions,1K 2K

From we get,5)0( =y

5221453.01 =K

221453.51 =K

Now

)cos221453.0sin415224.0()sin221453.0cos415224.0(

25.125.1 25.1225.1

2

25.1

1

xxexxe

eKxeKeKdx

dy

xx

xxx

++

+=

From ,3)0( =dx

dywe get

3)0cos(221453.0)0sin(415224.0())0sin(221453.0)0cos(415224.0(

)0(25.125.1

00

)0(25.1

2

)0(25.1

2

)0(25.1

1

=++

+

ee

eKeKeK

3415224.0221453.025.1 21 =++ KK 193771.325.1 21 =+ KK

193771.3)221453.5(25.1 2 =+ K

720582.92 =K

Substituting

and221453.51 =K

720582.92 =K

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08.01.14 Chapter 08.01

in the solution, we get

)cos221453.0sin415224.0()720582.9221453.5( 25.1 xxeexy xx ++=

The forms of the particular part of the solution for different right hand sides of ordinary

differential equations are given below

X ( )xyP 2

210 xaxaa ++ 2

210 xbxbb ++ axe axAe

)sin(bx )cos()sin( bxBbxA +

)sin(bxeax ( ))cos()sin( bxBbxAeax +

)cos(bx )cos()sin( bxBbxA +

)cos(bxe

ax

( ))cos()sin( bxBbxAeax +

Laplace Transforms

If is defined at all positive values of)(xfy = x , the Laplace transform denoted by is

given by

)(sY

(35)dxxfexfLsY sx )()}({)(0

==

where is a parameter, which can be a real or complex number. We can get back by

taking the inverse Laplace transform of .

s )(xf

)(sY

(36))()}({1 xfsYL =

Laplace transforms are very useful in solving differential equations. They give the solutiondirectly without the necessity of evaluating arbitrary constants separately.

The following are Laplace transforms of some elementary functions

sL

1)1( =

1

!)(

+=

n

n

s

nxL , where ....3,2,1,0=n

as

eL ax

=

1)(

22)(sin

as

aaxL

+=

22)(cos

as

saxL

+=

22)(sinh

as

aaxL

=

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22)(cosh

as

saxL

= (37)

The following are the inverse Laplace transforms of some common functions

111 =

s

L

axeas

L =

11

( )!11 11

=

n

x

sL

n

n, where ......3,2,1=n

( ) ( )!11 11

=

n

xe

asL

nax

n

axaas

L sin11

22

1 =

+

axas

sL cos22

1=

+

axaas

L sinh11

22

1 =

atas

sL cosh

22

1 =

( )bxe

bbasL ax sin

1122

1 =

+

( )bxe

bas

asL ax cos

22

1 =

+

( )axx

aas

sL sin

2

1222

1 =

+

(38)

Properties of Laplace transforms

Linear property

If are constants and and are functions ofcba ,, ),(),( xgxf )(xh x then

))(())(())(()]()()([ xhcLxgbLxfaLxchxbgxafL ++=++ (39)

Shifting property

If

(40))()}({ sYxfL =

then

(41))()}({ asYxfeL at =

Using shifting property we get

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08.01.16 Chapter 08.01

( )( ) 1

!+

=

n

nax

as

nxeL , 0n

( )( ) 22

sinbas

bbxeL ax

+=

( ) ( ) 22cos

basasbxeL ax+

=

( )( ) 22

sinhbas

bbxeL ax

=

( )( ) 22

coshbas

asbxeL ax

= (42)

Scaling property

If

(43))()}({ sYxfL =

then

=a

sY

aaxfL

1)}({ (44)

Laplace transforms of derivatives

If the first n derivatives of are continuous then)(xf

(45)

=0

)()}({ dxxfexfL nsxn

Using integration by parts we get

+

+++

=

0

001132

21

)()()1(

)()()1(......)()(

)()()()(

dxxfes

xfesxfes

xfesxfedxxfe

sxnn

sxnnnsx

nsxnsx

nsx

+=0

13221 )()0(.............)0()0()0( dxxfesfsfssff sxnnnnn

(46))0(........)0()0()0()( 13221 fsfssffsYs nnnnn =

Laplace transform technique to solve ordinary differential equations

The following are steps to solve ordinary differential equations using the Laplace transformmethod

(A) Take the Laplace transform of both sides of ordinary differential equations.(B) Express )(sY as a function ofs .(C) Take the inverse Laplace transform on both sides to get the solution.

Let us solve Examples 1 through 4 using the Laplace transform method.

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Example 6

Solve

xeydx

dy =+ 23 , 5)0( =y

Solution

Taking the Laplace transform of both sides, we get

( )xeLydx

dyL

=

+ 23

1

1)(2)]0()([3

+=+

ssYyssY

Using the initial condition, 5)0( =y we get

1

1)(2]5)([3

+=+

ssYssY

151

1)()23( ++

=+s

sYs

1

1615)()23(

+

+=+

s

ssYs

)23)(1(

1615)(

++

+=

ss

ssY

Writing the expression for in terms of partial fractions)(sY

231)23)(1(

1615

++

+=

++

+

s

B

s

A

ss

s

)23)(1(

23

)23)(1(

1615

++

+++

=++

+

ss

BBsAAs

ss

s

BBsAAss +++=+ 231615

Equating coefficients of and gives1s 0s

153 =+ BA 162 =+ BAThe solution to the above two simultaneous linear equations is

1=A 18=B

23

18

1

1)(

++

+

=

sssY

666667.06

11

++

+=

ss

Taking the inverse Laplace transform on both sides

++

+

=

666667.0

6

1

1)}({ 111

sL

sLsYL

Since

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08.01.18 Chapter 08.01

ateas

L =

+

11

The solution is given byxx eexy 666667.06)( +=

Example 7Solve

xeydx

dy 5.132 =+ , 5)0( =y

Solution

Taking the Laplace transform of both sides, we get

( )xeLydx

dyL 5.132

=

+

5.1

1)(3)]0()([2

+

=+

s

sYyssY

Using the initial condition , we get5)0( =y

5.1

1)(3]5)([2

+=+

ssYssY

105.1

1)()32( +

+=+

ssYs

5.1

1610)()32(

+

+=+

s

ssYs

)32)(5.1(

1610)(

++

+=

ss

ssY

)5.1)(5.1(21610++

+=ss

s

2)5.1(2

1610

+

+=

s

s

2)5.1(

85

+

+=

s

s


22 )5.1(5.1)5.1(

85

++

+=

+

+

s

B

s

A

s

s

22 )5.1(5.1

)5.1(85

+++=

++

sBAAs

ss

BAAss ++=+ 5.185

Equating coefficients of and gives1s 0s 5=A

85.1 =+ BAThe solution to the above two simultaneous linear equations is

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5=A5.0=B

2)5.1(

5.0

5.1

5)(

++

+=

sssY


++

+=

2

111

)5.1(

5.0

5.1

5)}({

sL

sLsYL

Since

axeas

L =

+

11and axxe

asL =

+ 21

)(

1

The solution is given byxx xeexy 5.15.1 5.05)( +=

Example 8

Solve

xydx

dy

dx

ydsin125.332

2

2

=++ , 3)0(,5)0( === xdx

dyy

Solution

Taking the Laplace transform of both sides

( )xLydx

dy

dx

ydL sin125.332

2

2

=

++

and knowing

2

2

dxydL ( ) ( ) ( )002 == x

dxdysysYs

dx

dyL ( ) ( )0yssY =

1

1)(sin

2 +=

sxL

we get

[ ]1

1)(125.3)0()(3)0()0()(2

2

2

+=++

=

ssYyssYx

dx

dysysYs

[ ] [ ] 11)(125.35)(335)(2

22

+=++ ssYssYssYs

( )[ ]1

12110)(125.332

2 +=++

sssYss

( )[ ] 21101

1)(125.332

2++

+=++ s

ssYss

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08.01.20 Chapter 08.01

[ ])1(

21101022)(125.332

2

232

+

+++=++

s

ssssYss

( )( )125.332122102110

)(22

23

+++

+++=

sss

ssssY


( ) ( ) ( )( )125.3321

22102110

1125.332 22

23

22 +++

+++=

+

++

++

+

sss

sss

s

DCs

ss

BAs

( )( )

( )( )125.332122102110

1125.332

125.332125.332

22

23

22

22323

+++

+++=

+++

+++++++++

sss

sss

sss

DDsDsCsCsCsBBsAsAs

( ) ( ) ( ) ( )( )( )

( )( )125.332122102110

125.3321

125.33125.3232

22

23

22

23

+++

+++=

+++

+++++++++

sss

sss

sss

DBsDCAsDCBsCA

Equating terms of , and gives3s 12 , ss 0s

102 =+ CA2123 =++ DCB

103125.3 =++ DCA22125.3 =+ DB

The solution to the above four simultaneous linear equations is584474.10=A657534.21=B292237.0=C

109589.0=DHence

1

109589.0292237.0

125.332

657534.21584474.10)(

22 +

++

++

+=

s

s

ss

ssY

( ) }1)75.0{(2}1)5625.05.1{(2125.332 222 ++=+++=++ sssss

1

109589.0292237.0

}1)75.0{(2

719179.13)75.0(584474.10)(

22 +

++

++

++=

s

s

s

ssY

)1(

109589.0

)1(

292237.0

}1)75.0{(

859589.6

}1)75.0{(

)75.0(292237.5

2222 ++

+

+++

++

+=

ss

s

ss

s

Taking the inverse Laplace transform of both sides

++

+

+++

++

+=

1

109589.0

1

292237.0

1)75.0{(

859589.6

}1)75.0{(

)75.0(292237.5)}({

2

1

2

1

2

1

2

11

sL

s

sL

sL

s

sLsYL

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++

+

+++

++

+=

1

1109589.0

1292237.0

1)75.0{(

1859589.6

}1)75.0{(

75.0292237.5)}({

2

1

2

1

2

1

2

11

sL

s

sL

sL

s

sLsYL

Since

( )bxe

bas

asL ax cos

22

1 =

++

+

( )bxe

bas

bL ax sin

22

1 =

++

axas

L sin1

22

1 =

+

axas

sL cos

22

1 =

+


xx

xexexyxx

sin109589.0cos292237.0

sin8595859.6cos292237.5)( 75.075.0

+

+=

( ) xxxxe x sin109589.0cos292237.0sin859589.6cos292237.5 75.0 ++=

Example 9

Solve

xydx

dy

dx

ydcos125.362

2

2

=++ , 3)0(,5)0( === xdx

dyy

SolutionTaking the Laplace transform of both sides

( )xLydx

dy

dx

ydL cos125.362

2

2

=

++

and knowing

2

2

dx

ydL ( ) ( ) ( )002 == x

dx

dysysYs

dx

dyL ( ) ( )0yssY =

1)(cos

2 +=

s

sxL

we get

[ ]1

)(125.3)0()(6)0()0()(22

2

+=++

=

s

ssYyssYx

dx

dysysYs

[ ] [ ]1

)(125.35)(635)(22

2

+=++

s

ssYssYssYs

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08.01.22 Chapter 08.01

[ ] 36101

)(125.3)62(2

+++

=++ ss

ssYss

[ ]1

36111036)(125.362

2

232

+

+++=++

s

ssssYss

( )( )125.362136113610

)( 22

23

+++

+++= sss

ssssY


( ) ( ) ( )( )125.3621

36113610

1125.36222

23

22 +++

+++=

+

++

++

+

sss

sss

s

DCs

ss

BAs

( )( )

( )( )125.362136113610

1125.362

125.362125.362

22

23

22

22323

+++

+++=

+++

+++++++++

sss

sss

sss

DDsDsCsCsCsBBsAsAs

( ) ( ) ( ) ( )( )( )

( )( )125.362136113610

125.3621125.36125.3262

22

23

22

23

+++

+++=

++++++++++++

sss

sss

sssDBsDCAsDCBsCA

Equating terms of , and gives3s12 , ss 0s

102 =+ CA3626 =++ DCB

116125.3 =++ DCA36125.3 =+ DB

The solution to the above four simultaneous linear equations is939622.9=A496855.35=B

0301886.0=C

161006.0=DThen

1

161006.00301886.0

125.362

496855.35939622.9)(

22 +

++

++

+=

s

s

ss

ssY

( ) }829156.0)5.1{(2}6875.0)25.23{(2125.362 2222 +=++=++ sssss

1

161006.00301886.0

}829156.0)5.1{(2

587422.20)5.1(939622.9)(

222 +

++

+

++=

s

s

s

ssY

1

161006.0

1

0301886.0

}829156.0)5.1{(

293711.10

}829156.0)5.1{(

)5.1(969811.4

22

2222

++

++

++

++=

ss

s

ss

s


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++

++

++

+

+=

1

161006.0

1

0301886.0

829156.0)5.1{(

293711.10

}829156.0)5.1{(

)5.1(969811.4)}({

2

1

2

1

22

1

22

11

sL

s

sL

sL

s

sLsYL

++

++=

22

1

22

1

829156.0)5.1(1293711.10

829156.0)5.1()5.1(969811.4

sL

ssL

++

++

)1(

1161006.0

)1(0301886.0

2

1

2

1

sL

s

sL

Since

( )bxe

bas

asL ax cosh

22

1 =

+

+

( )bxe

bbasL ax sinh

1122

1 =

+

axaas

L sin11

22

1 =

+

axas

sL cos

22

1 =

+


xexexy xx

sin161006.0cos0301886.0

)829156.0sinh(829156.0

293711.10)829156.0cosh(969811.4)( 5.15.1

++

+=

eeee

e

xxxxx

sin161006.0cos030188.0

2414685.122969811.4

829156.0829156.0829156.0829156.05.1

++

+

+

=

( )x

xeee xxx

sin161006.0

cos0301886.0722437.3692248.8 829156.0829156.05.1

+

+=

Example 10

Solve

xey

dx

dy

dx

yd x sin125.3522

2=++ , 3)0(,5)0( === x

dx

dyy

Solution

Taking the Laplace transform of both sides

( )xeLydx

dy

dx

ydL x sin125.352

2

2=

++

knowing

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08.01.24 Chapter 08.01

2

2

dx

ydL ( ) ( ) ( )002 == x

dx

dysysYs

dx

dyL ( ) ( )0yssY =

1)1(1)sin(

2 ++=

sxeL x

we get

[ ]

[ ] [ ]1)1(

1)(125.35)(535)(2

1)1(

1)(125.3)0()(5)0()0()(2

2

2

2

2

++=++

++=++

=

ssYssYssYs

ssYyssYx

dx

dysysYs

( )[ ]1)1(

13110)(125.352

2 ++=++

sssYss

[ ] 31101)1(

1)(125.3)52(2 ++++

=++ ss

sYss

[ ]22

51821063)(125.352

2

232

++

+++=++

ss

ssssYss

( )( )125.3522263825110

)(22

23

++++

+++=

ssss

ssssY


( )( )125.35222

63825110

22125.35222

23

22

++++

+++=

++

++

++

+

ssss

sss

ss

DCs

ss

BAs

( )( )

( )( )125.3522263825110

22125.352

2222125.352125.352

22

23

22

223223

++++

+++=

++++

+++++++++++

ssss

sss

ssss

BBsBsAsAsAsDDsDsCsCsCs

( ) ( ) ( ) ( )( )( )

( )( )125.3522263825110

125.35222

2125.3225125.32252

22

23

22

23

++++

+++=

++++

+++++++++++

ssss

sss

ssss

BDsBADCsBADCsAC

Equating terms of , and gives four simultaneous linear equations3s12 , ss 0s

102 =+ AC51225 =+++ BADC

82225125.3 =+++ BADC 632125.3 =+ BDThe solution to the above four simultaneous linear equations is

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442906.10=A

494809.32=B

221453.0=C

636678.0=DThen

22636678.0221453.0

125.352494809.32442906.10)(

22 +++

+++=

sss

ssssY

( ) 222 )25.1(2)}5625.15.2{(2125.352 +=++=++ sssss

1)1(

415225.0)1(221453.0

)25.1(2

441176.19)25.1(442906.10)(

22 ++

++

+

++=

s

s

s

ssY

1)1(

415225.0

1)1(

)1(221453.0

)25.1(

720588.9

)25.1(

)25.1(221453.5

2222 ++

++

+

++

+

+=

ss

s

ss

s


++

++

+

++

+=

1)1(

415225.0

1)1(

)1(221453.0

)25.1(

720588.9

)25.1(

221453.5)}({

2

1

2

1

2

111

sL

s

sL

s

L

s

LsYL

++

++

+

++

+=

1)1(

1415225.0

1)1(

)1(221453.0

)25.1(

1720588.9

)25.1(

1221453.5

2

1

2

1

2

11

sL

s

sL

sL

sL

Since

( ) bxebasas

Lax

cos221

=

++

+

( )bxe

bas

bL ax sin

22

1 =

++

axeas

L =

+

11

)!1()(

1 11

=

+

n

xe

asL

nax

n


xexexeexy

x

xxx

sin415225.0cos221453.0720588.9221453.5)(

25.125.1

+=

( ) )sin415225.0cos221453.0(720588.9221453.5 25.1 xxexe xx ++=

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08.01.26 Chapter 08.01

ORDINARY DIFFERENTIAL EQUATIONS

Topic A Primer on ordinary differential equationsSummary Textbook notes of a primer on solution of ordinary differential equations

Major All majors of engineering

Authors Autar Kaw, Praveen ChalasaniDate April 24, 2009

Web Site http://numericalmethods.eng.usf.edu
http://../http://../

ode primer

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